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Georgia [21]
2 years ago
13

I need someone to help me work through this

Physics
1 answer:
Sergio039 [100]2 years ago
3 0
The average rate of change of distance over the time interval 
3 ≤ t ≤ 6 represents the coin's average velocity over that interval.
You might be interested in
The frequency of a wave is 200 Hz. The wavelength is 0.1 m. What is the period of the wave?
Aleks04 [339]
The formula for the period of wave is: wave period is equals to 1 over the frequency.waveperiod=\frac{1}{frequency}
To get the value of period of wave you need to divide 1 by 200 Hz. However, beforehand, you have to convert 200 Hz to cycles per second. So that would be, 200 cyles per second or 200/s.
By then, you can start the computation by dividing 1 by 200/s. Since 200/s is in fractional form, you have to find its reciprocal form and multiply it to one which would give you 1 (one) second over 200. This would then lead us to the value 0.005 seconds as the wave period.

wave period= 1/200 Hz
Convert Hz to cycles per second first
200 Hz x 1/s= 200/second
Make 200/second as your divisor, so:

wave period= 1/ 200/s

get the reciprocal form of 200/s which is s/200

then you can start the actual computation:

wave period= 1 x s divided by 200

this would give us an answer of 0.005 s. 
6 0
2 years ago
A satellite orbits Earth. The only force on the satellite is the gravitational force exerted by Earth. How does the satellite’s
bulgar [2K]

Answer:

here we can say that acceleration of the satellite is same as the gravitational field due to Earth at that location

Explanation:

As we know that gravitational field is defined as the force experienced by the satellite per unit of mass

so we will have

E = \frac{F}{m}

now in order to find the acceleration of the satellite we know by Newton's II law

F = ma

so we will have

a = \frac{F}{m}

so here we can say that acceleration of the satellite is same as the gravitational field due to Earth at that location

4 0
3 years ago
Read 2 more answers
A mass m is attached to an ideal massless spring. When this system is set in motion with amplitude a, it has a period t. What is
Luden [163]

The period will be the same if the amplitude of the motion is increased to 2a

What is an Amplitude?

Amplitude refers to the maximum extent of a vibration or oscillation, measured from the position of equilibrium.

Here,

mass m is attached to the spring.

mass attached = m

time period = t

We know that,

The time period for the spring is calculated with the equation:

T = 2\pi \sqrt{\frac{m}{k} }

Where k is the spring constant

Now if the amplitude is doubled, it means that the distance from the equilibrium position to the displacement is doubled.

From the equation, we can say,

Time period of the spring is independent of the amplitude.

Hence,

Increasing the amplitude does not affect the period of the mass and spring system.

Learn more about time period here:

<u>brainly.com/question/13834772</u>

#SPJ4

7 0
2 years ago
An example of a high explosive is what?
nordsb [41]
The answer is Dynamite.
Explosive, any substance or device that can be made to produce a volume of rapidly expanding gas in an extremely brief period. Chemical explosives are of two types; detonating, or high explosives and deflagrating, or low, explosives. Detonating explosives, such as TNT and dynamite, are characterized by extremely rapid decomposition and development of high pressure, whereas deflagrating explosives, such as black and smokeless powders, involve merely fast burning and produce relatively low pressures.
5 0
3 years ago
Read 2 more answers
A projectile is fired horizontally from a height of 78.4 m at a speed of 300 m/sec. How far did it travel horizontally before hi
Mice21 [21]

Answer:

Explanation:

Using the formula for calculating range expressed as;

R = U√2H/g

U is the speed = 300m/s

H is the maximum height = 78.4m

g is the acceleration due to gravity = 9.8m/s²

Substitute into the fromula;

R = 300√2(78.4)/9.8

R = 300 √(16)

R = 300*4

R = 1200m

Hence the projectile travelled 1200m before hitting the ground

3 0
3 years ago
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